Chapter 11Beam on Elastic Foundation

11.1 Overview

In some applications such as rail tracks, the member subjected to loads is
supported on continuous foundations. That is the reactions due to external
loading is distributed along the length of the member. Here we study on how to
get the stresses and displacements in these members resting on continuous
foundations. If the dimensions of this member is such that, it is longer
along one of the axis, called the longitudinal axis in comparison with
the dimensions along the other directions, it is called as a beam. If we
assume that the reaction force offered by the continuous support is a
function of the displacement that of the member, the support is called as
elastic. A beam resting on an elastic support is said to be beam on elastic
foundation.

In this chapter, we first formulate this problem of beam on an elastic
foundation for a general loading condition. Then, we study the problem of a
concentrated load at the mid point of a beam that is infinitely long. Appealing to
the principle of superposition we obtain the solution to the problem of a
concentrated moment at mid span and uniformly distributed load of length L,
centered about the midpoint of the beam.

11.2 General formulation

Figure 11.1: Schematic of a long beam on elastic foundation

In this section, we formulate the boundary value problem of beam on an
elastic foundation. A beam having some cross section, resting on an elastic
support is shown in figure 11.1. We assume that the reaction offered by the
support at any point is directly proportional to the displacement of that point
along the y direction and is in a direction opposite to the displacement. Thus, if Δ
is the vertical displacement of a point in the beam, qy the support reaction per
unit width of the beam, then the above assumption that the reaction
force is proportional to the displacement mathematically translates into
requiring

(11.1)

Assuming the beam to be homogeneous, we obtained the equation (8.41)
which we document here again:

(11.2)

where yo is the y coordinate of the centroid of the cross section which
can be taken as 0 without loss of generality provided the origin of the
coordinate system used is located at the centroid of the cross section, E is the
Young’s modulus, (x,y) is the coordinate of the point along the axis of
the beam direction and the y direction, Mz is the z component of the
bending moment, Izz is the moment of inertia of the section about the z
axis.

In section 8.1, we integrated the equilibrium equations and obtained equations
(8.18) and (8.25) which we record here:

where Vy is the shear force along the y direction and qy is the transverse
loading along the y direction. Combining the equations (11.3) and (11.4) we
obtain,

where we have assumed that the origin is located at the centroid of the
cross section and hence have set yo = 0.

11.3 Example 1: Point load

Figure 11.2: Schematic of a long beam on elastic foundation subjected to
concentrated load at mid span

The first boundary value problem that we study for the beam on elastic
foundation is when it is subjected to a point load at its mid span as shown in
figure 11.2. The origin of the coordinate system is assumed to coincide with the
point of application of the load. The beam length is assumed to be large enough
compared to its lateral dimension that it can be considered to be infinitely long.
(We shall quantify what length could be considered as infinitely long after we
obtain the solution.)

Figure 11.3: Free body diagram of half the section of long beam on elastic
foundation subjected to concentrated load

To obtain the solution we section the beam at x = 0, the point of
application of the concentrated load, as shown in figure 11.3. Since, there is a
concentrated force acting at x = 0, the shear force would be discontinuous at x =
0 and hence the fourth derivative of the deflection, Δ does not exist.
Consequently, the governing equation (11.9) is valid only in the domain
x > 0 and x < 0 and not at x = 0. Therefore, we segment the beam at
x = 0 and solve (11.9) on each of the segments. Then, we ensure, the
differentiability of the second order derivative of deflection so that the third order
derivative exist. This is required to ensure the existence of shear force at x =
0.

We also expect the deflection to be symmetric about x = 0 that is, Δ(x) =
Δ(-x) and therefore the slope of the deflection should be zero at x = 0,
i.e.,

(11.12)

By sectioning the beam at x = 0 we find the bending moment and shear force at
this location. Using equation (11.2) we find that

(11.13)

where Mo is the bending moment at x = 0, acting as shown in the figure 11.3 and
the negative sign is to account for the fact that it is hogging. Since, shear force
should exist, the continuity of the bending moment and has to be ensured.
Therefore the value of bending moment at both the segments of the beam should
be the same.

Substituting equation (11.2) in (11.3) and assuming the beam to be
homogeneous and prismatic, we obtain

since, we expect the effect of the load would be felt only in its vicinity.

To obtain the solution, we first focus on the right half of the beam wherein x> 0. Then, the requirement (11.17) implies that the constants C3 and C4 in the
general solution (11.10) has to be zero; otherwise Δ →∞ as x →∞. Next, the
condition (11.12) requires that C1 = C2 = C0. Finally, the equation (11.15) tells
us that

(11.18)

where to obtain the last equality we have made use of (11.8). Thus, in the domain,
x > 0,

(11.19)

Now, we consider the left half of the beam, i.e., x < 0. The requirement
(11.17) implies that the constants C1 and C2 in the general solution (11.10) has to
be zero. Next, the condition (11.12) requires that C3 = -C4 = C5. Finally, the
equation (11.16) tells us that

(11.20)

where to obtain the last equality we have made use of (11.8). Thus, in the domain,
x < 0,

(11.21)

Thus, the distribution of the reaction from the foundation along the axis of the
beam given in (11.1) evaluates to:

(11.22)

The variation of the bending moment along the axis of the beam obtained from
(11.2) is:

(11.23)

The shear force variation along the axis of the beam computed using (11.14)
is:

(11.24)

(a) Variation of support reaction(b) Variation of bending moment(c) Variation of shear force

Figure 11.4: Variation of support reaction, bending moment and shear force
along the axis of the beam on elastic foundation

In figure 11.4 we plot the variation of the support reaction, bending moment
and shear force along the axis of the beam. It can be seen from the figure
that though the beam is assumed to be infinitely long the reaction force,
bending moment and shear forces all tend to zero for βx > 5. Hence,
a beam may be considered as long if its length is greater than, 5∕β. It
can also be seen from the figure that the maximum deflection, support
reaction, bending moment and shear occurs at z = 0 and these values
are,

(11.25)

It can be seen from figure 11.4a that the support reaction changes sign. The
support reaction changes sign at a point when qy = 0, i.e., sin(βx) = - cos(βx) or
at x = 3π∕(4β). Since, the support reaction is proportional to the deflection, Δ,
this change in sign of the support reaction also tells us that the beam will uplift at
x = ±3π∕(4β). Hence, the beams have to be adequately clamped to the
foundation to prevent it from uplifting.

11.4 Example 2: Concentrated moment

Figure 11.5: Schematic of a long beam on elastic foundation subjected to
concentrated moment at mid span

The concentrated moment, Mo, is considered to be equivalent to the
action of two concentrated forces, P, equal in magnitude but opposite
in direction and separated by a distance L as shown in the figure 11.5.
Thus,

(11.26)

We obtain the solution to this loading case by superposing the displacement
field obtained in the above example for a single point load. Thus, it follows from
equations (11.17) and (11.19) that the displacement due to the downward acting
force at a distance, L∕2 from the origin is,

(11.27)

Similarly, the displacement due to the upward acting force at a distance, -L∕2
from the origin is

(11.28)

Since, the displacement is small and the material obeys Hooke’s law,
we can superpose the solutions as discussed in section 7.5.2. Hence, the
displacement under the action of both the forces, Δ = ΔL∕2 + Δ-L∕2 evaluates
to,

(11.29)

where we have used equation (11.26). When L → 0 and PL → Mo the above
equation (11.29) evaluates to,

(11.30)

Having found the displacement, (11.30), the variation of the bending moment
along the axis of the beam obtained from (11.2) is:

11.5 Example 3: Uniformly distributed load

Figure 11.6: Schematic of a long beam on elastic foundation subjected to
uniformly distributed load of length L on either side of the mid span

Next, we study the problem of an infinite beam on an elastic foundation
subjected to a uniformly distributed load of length L symmetrically on either side
of the origin, as shown in figure 11.6. As before, we apply the principle of
superposition to find the deflection at a point to be

11.6 Summary

In this chapter, we formulated and solved the problem of a concentrated load
acting on a long beam on elastic foundation. Using this solution and appealing to
the principal of superposition, we solved two problems. One of the problems is
that of a concentrated moment on a long beam on elastic support. The other
problem is that of uniformly distributed load of length L, on a long beam
continuously supported at the bottom. These problem serve as an illustration of
the use of principle of superposition.

11.7 Self-Evaluation

A steel beam of a rectangular cross section, 180 mm wide and 280
mm thick, is resting on an elastic foundation whose modulus of
foundation is 6.5 N/mm2. This beam is subjected to a concentrated
anti-clockwise moment of 0.5 MNm at the center. Determine the maximum
deflection and the maximum bending stresses in the beam. Assume
Young’s modulus, E = 200 GPa and the Poisson’s ratio, ν = 0.3. Also,
plot

The deflected shape of the beam

The variation in the bending moment along the axis of the beam

The variation in the shear force along the axis of the beam

Find the length of the beam beyond which it would require clamping to
prevent uplift.

Repeat problem - 1, for the case in which the beam is subjected to a
uniformly distributed load of intensity 200 N/mm over a length 500 mm
about the center, instead of a concentrated moment.

A four-wheel wagon runs on steel rails. The rails have a depth of 140mm;
the distance from the top of a rail to its centroid is 70 mm; and its
moment of inertia is 21 * 106mm4. The Young’s modulus of the rail, E
= 210 GPa and the Poisson’s ratio, ν = 0.3. The rail rests on an
elastic foundation with linear spring constant, Ks = 12N∕mm2.
The two wheels on each side of the car are spaced 2.50 m center to
center and the distance between the front and rear axle is 13 m. If
each wheel load is 90 kN, determine the maximum deflection and
maximum bending stress on the rail. List the assumptions made in the
analysis.

Determine the thickness of a square foundation of side 1 m required, if it
were to carry a concentrated load of 1 MN and a concentrated clockwise
moment of 0.1 MNm. Assume that the foundation rests on an elastic
foundation with spring constant, Ks = 12N∕mm2, the Young’s modulus of
the foundation is 20 GPa and that the maximum permissible normal stresses
on the foundation is 10 MPa.